What is the slope of the line tangent to $f(x) = -2x^{2}-4x+4$ at $x = 1$ ?
Solution: The slope of the tangent line is $ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$ $ = \lim_{h \to 0} \frac{(-2(x+h)^{2}-4(x+h)+4) - (-2x^{2}-4x+4)}{h}$ $ = \lim_{h \to 0} \frac{(-2(x^{2}+2x h+h^{2})-4(x+h)+4) - (-2x^{2}-4x+4)}{h}$ $ = \lim_{h \to 0} \frac{-2x^{2}-4(x h)-2h^{2}-4x-4h+4+2x^{2}+4x-4}{h}$ $ = \lim_{h \to 0} \frac{-4(x h)-2h^{2}-4h}{h}$ $ = \lim_{h \to 0} -4x-2h-4$ $ = -4x-4$ $ = (-4)(1)-4$ $ = -8$